of 1 35
Proton-Seconds And The Vector Equilibrium
Ian Beardsley!
April 19, 2022"
of 2 35
Abstract: We observe six-fold symmetry in space and time and that through Buckminster
Fuller’s Vector Equilibrium we see the periodic table of the elements is an extension of
space and time. The unit of a second follows from this, as well as the hydrocarbons, the
skeletons of life chemistry, the radius of a proton, the radius of the solar system, and the
relative abundances of the most abundant elements in the universe hydrogen and helium.
We further suggest the crux of the periodic table is in that the life elements describe the
artificial intelligence elements.!
of 3 35
Contents
1.0 Space and Time……………………………………………4
2.0 The Periodic Table of the Elements…………………….8
3.0 Proton-Seconds And The Vector Equilibrium………..12
4.0 Formulation of Proton-Seconds………………………..32
of 4 35
1.0 Space And Time
We live in a 3 dimensional space. It requires 3 values to
specify a point in space such as the point (1, 1, 1).!
!
We can only draw the 3 dimensional shadow of a four
dimensional hypercube because we cannot find a fourth right
angle in our mind or space.!
!
But we can draw a third right angle to a plane for which we
see is the 2 dimensional shadow of a three dimensional cube.!
of 5 35
And as we move through a distance x we see
we see we have travelled through a time t.!
In order to specify a point P in space we need
three values See illustration on page 7:!
!
And, in order to do operations on how these
three values may vary we need three unit
vectors to specify their directions, each
specified by three values two distinctly 0:!
!
!
!
And three distinct operators (such as , ,
. We define (nabla) and call it the del
operator:!
!
If we have a vector field , Then, where in a two dimensional plane we had one kind of
multiplication, in three dimensions we we have two kinds of multiplication the dot product and
the cross product:!
!
Is a scalar. This, the dot product is also called the scalar product.!
!
Is a vector.!
"
(x, y, z) = (a, b, c)
i = (1,0,0)
j = (0,1,0)
k = (0,0,1)
∂
∂x
∂
∂y
∂
∂z
∇
∇ =
i
∂
∂x
+
j
∂
∂y
+
k
∂
∂z
A
∇ ⋅
A =
∂
A
x
∂x
+
∂
A
y
∂y
+
∂
A
z
∂z
∇ ×
A =
i
j
k
∂
∂x
∂y
∂y
∂
∂z
A
x
A
y
A
z
of 6 35
We don’t have to cross nabla with , but we can cross with another vector as well dot
them:!
!
!
!
Or we can cross with :!
=!
!
In this instance we have three components each consisting of two terms for a total of six
distinct terms, or products. They form a three by three matrix is elements. The key point
I would like to make here is that there are six terms as there are six faces on a cube, and 2
times 9 is 18 is 6 times 3. The periodic table of the elements is periodic over 18 groups, each
group of elements having related properties like number of valence electrons. For example
carbon is in group 14 and has 18-14=4 valence electrons and silicon is in group 14 as well,
below it in period 3 with 18-14=4 valence electrons as well. Regardless of whether we are
working in rectangular coordinates as we have done so far, or some other coordinate system,
like cylindrical or spherical, for which we need three values to specify a point as well. Thus we
call it . Where is the real numbers. However, we see cylindrical coordinates, which are
employed for doing computations about straight lines like for the electric field that arises due to
electrons moving in a wire, because as we progress along the straight line all distances from it
are the same, the radius r of a cylinder, that it is contained int the rectangular coordinates
because r is measured from the x-axis, is measured from the x-axis, and z is measured up
from the x-y plane. Similarly, spherical coordinates are used for doing computations with
spherical symmetry, like for the electric charge that arises from a charge because the electric
field arises radially in all directions equally, so r is the same in all directions as well, and
spherical coordinates are as well contained in the rectangular coordinate system because r is
measured from the intersection of the intersection of the x,y, and z axes. is measured from
the x-axis and is measured from the z-axis. We see in fact that (See illustration next page,
):!
Cylindrical Coordinates!
, , z=z, , !
Spherical coordinates!
, , , "
A
A
B
A = A
1
i + A
2
j + A
3
k
B = B
1
i + B
2
j + B
3
k
A ⋅
B = A
1
B
1
+ A
2
B
2
+ A
3
B
3
A
B
A ×
B =
i
j
k
A
1
A
2
A
3
B
1
B
2
B
3
(A
2
B
3
− A
3
B
2
)
i − (A
3
B
1
− A
1
B
3
)
j + (A
1
B
2
− A
2
B
1
)
k
3
2
= 9
ℝ
3
ℝ
θ
θ
ϕ
ℝ
3
x = rcosθ
y = rsinθ
r
2
= x
2
+ y
2
ta nθ = y/x
x = rsinϕcosθ
y = rsinϕsinθ
z = rcosϕ
r
2
= x
2
+ y
2
+ z
2
of 7 35
"
of 8 35
2.0 The Periodic Table of The Elements
We wish to establish a theorem from all that has been said here.!
Theorem 1.0 The Periodic Table of the Elements employs six fold symmetry. The periods of the
periodic table are periodic over 18 groups, which is 3 times 6, the three being the components
of , the number of values needed to specify a point in such a space.!
Above we see the artificial intelligence (AI) elements pulled out of the periodic table of the elements. As
you see we can make a 3 by 3 matrix of them and an AI periodic table. Silicon and germanium are in
group 14 meaning they have 4 valence electrons and want 4 for more to attain noble gas electron
configuration. If we dope Si with B from group 13 it gets three of the four electrons and thus has a
deficiency becoming positive type silicon and thus conducts. If we dope the Si with P from group 15 it
has an extra electron and thus conducts as well. If we join the two types of silicon we have a
semiconductor for making diodes and transistors from which we can make logic circuits for AI.
As you can see doping agents As and Ga are on either side of Ge, and doping agent P is to the right of Si
but doping agent B is not directly to the left, aluminum Al is. This becomes important. I call (As-Ga) the
differential across Ge, and (P-Al) the differential across Si and call Al a dummy in the differential because
boron B is actually used to make positive type silicon.
That the AI elements make a three by three matrix they can be organized with the letter E with subscripts
that tell what element it is and it properties, I have done this:
Thus E24 is in the second row and has 4 valence electrons making it silicon (Si), E14 is in the first row
and has 4 valence electrons making it carbon (C). I believe that the AI elements can be organized in a 3 by
3 matrix makes them pivotal to structure in the Universe because we live in three dimensional space so
the mechanics of the realm we experience are described by such a matrix, for example the cross product.
Hence this paper where I show AI and biological life are mathematical constructs and described in terms
of one another.
We see, if we include the two biological elements in the matrix (E14) and and (E15) which are carbon and
nitrogen respectively, there is every reason to proceed with this reasoning if the idea is to show not only
are the AI elements and biological elements mathematical constructs, they are described in terms of one
another. We see this because the first row is ( B, C, N) and these happen to be the only elements that are
ℝ
3
E
13
E
14
E
15
E
23
E
24
E
25
E
33
E
34
E
35
of 9 35
not core AI elements in the matrix, except boron (B) which is out of place, and aluminum (Al) as we will
see if a dummy representative makes for a mathematical construct, the harmonic mean. Which means we
have proved our case because the first row if we take the cross product between the second and third rows
are, its respective unit vectors for the components meaning they describe them:
And silicon (Si) is at the center of our AI periodic table of the elements. We see the biological elements C
and N being the unit vectors are multiplied by the AI elements, meaning they describe them! But we have
to ask; Why does the first row have boron in it which is not a core biological element, but is a core AI
element? The answer is that boron is the one AI element that is out of place, that is, aluminum is in its
place. But we see this has a dynamic function.
The primary elements of artificial intelligence (AI) used to make diodes and transistors, silicon (Si) and
germanium (Ge) doped with boron (B) and phosphorus (P) or gallium (Ga) and arsenic (As) have an
asymmetry due to boron. Silicon and germanium are in group 14 like carbon (C) and as such have 4
valence electrons. Thus to have positive type silicon and germanium, they need doping agents from group
13 (three valence electrons) like boron and gallium, and to have negative type silicon and germanium they
need doping agents from group 15 like phosphorus and arsenic. But where gallium and arsenic are in the
same period as germanium, boron is in a different period than silicon (period 2) while phosphorus is not
(period 3). Thus aluminum (Al) is in boron’s place. This results in an interesting equation.
Equation 2.1
A = (Al, Si, P )
B = (G a, Ge, A s)
A ×
B =
B
C
N
Al Si P
G a Ge As
= (Si ⋅ As − P ⋅ G e)
B + (P ⋅ G a − Al ⋅ A s)
C + (Al ⋅ G e − Si ⋅ G a)
N
A ×
B = − 145
B + 138
C + 1.3924
N
A = 26.98
2
+ 28.09
2
+ 30.97
2
= 50g/m ol
B = 69.72
2
+ 72.64
2
+ 74.92
2
= 126g/m ol
A ⋅
B = A Bcosθ
cosθ =
6241
6300
= 0.99
θ = 8
∘
A ×
B = A Bsin θ = (50)(126)sin8
∘
= 877.79
877.79 = 29.6g /m ol ≈ Si = 28.09g /m ol
Si(A s − G a) + G e(P − Al )
SiG e
=
2B
Ge + Si
of 10 35
The differential across germanium crossed with silicon plus the differential across silicon crossed with
germanium normalized by the product between silicon and germanium is equal to the boron divided by
the average between the germanium and the silicon. The equation has nearly 100% accuracy (note: using
an older value for Ge here, it is now 72.64 but that makes the equation have a higher accuracy):
Due to an asymmetry in the periodic table of the elements due to boron we have the harmonic
mean between the semiconductor elements (by molar mass):
Equation 2.2.
This is Stokes Theorem if we approximate the harmonic mean with the arithmetic mean:
We can make this into two integrals:
Equation 2.3
Equation 2.4
If in the equation (The accurate harmonic mean form):
28.09(74.92 − 69.72) + 72.61(30.97 − 26.98)
(28.09)(72.61)
=
2(10.81)
(72.61 + 28.09)
0.213658912 = 0.21469712
0.213658912
0.21469712
= 0.995
Si
B
(As − G a) +
Ge
B
(P − Al ) =
2SiG e
Si + G e
∫
S
( ∇ × u ) ⋅ d S =
∮
C
u ⋅ d r
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
d x d y ≈
1
Ge − Si
∫
Ge
Si
x d x
∫
1
0
∫
1
0
Si
B
(As − Ga)d yd z ≈
1
3
1
(Ge − Si )
∫
Ge
Si
x d x
∫
1
0
∫
1
0
Ge
B
(P − Al )d x dz ≈
2
3
1
(Ge − Si )
∫
Ge
Si
yd y
of 11 35
Equation 2.5
We make the approximation
Equation 26.
Then the Stokes form of the equation becomes
Equation 2.7
Thus we see for this approximation there are two integrals as well:
Equation 2.8
Equation 2.9
Si
B
(As − Ga) +
Ge
B
(P − Al ) =
Ge − Si
Ge
Si
dx
x
2SiGe
Si + Ge
≈ G e − Si
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
d yd z =
∫
Ge
Si
d x
∫
1
0
∫
1
0
Si
B
(As − Ga)d yd z =
1
3
∫
Ge
Si
dz
∫
1
0
∫
1
0
Ge
B
(P − Al )dyd z =
2
3
∫
Ge
Si
dz
of 12 35
3.0 Proton-Seconds And The Vector Equilibrium
In order to present the elements as mathematical structures we need to explain the matter from
which they are made as mathematical constructs. We need a theory for Inertia. I had found
(Beardsley Essays In Cosmic Archaeology. 2021) where I suggested the idea of proton
seconds, that is six proton-seconds, which is carbon the core element of biological life if we
can figure out a reason to divide out the seconds. I found!
Equation 3.1 !
Where h is Planck’s constant 6.62607E—34 Js, is the radius of a proton 0.833E-15m, G is
the universal constant of gravitation 6.67408E-11 (Nm2)/(kg2), and c is the speed of light
299,792,459 m/s. And is t=1 second. is the Sommerfeld constant (or fine structure
constant) is 1/137. The mass of a proton is .!
The fine structure constant squared is the ratio of the potential energy of an electron in the first
circular orbit to the energy given by the mass of an electron in the Bohr model times the speed
of light squared: !
!
Matter is that which has inertia. This means it resists change in position with a force applied to
it. The more of it, the more it resists a force. We understand this from experience, but what is
matter that it has inertia?!
I would like to answer this by considering matter in one of its simplest manifestations, the
proton, a small sphere with a mass of 1.6726E-27 kg. This is a measure of its inertia.!
I would like to suggest that matter, often a collection of these protons, is the three dimensional
cross-section of a four dimensional hypersphere.!
The way to visualize this is to take space as a two-dimensional plane and the proton as a two
dimensional cross-section of a sphere, which would be a circle.!
In this analogy we are suggesting a proton is a three dimensional bubble embedded in a two
dimensional plane. As such there has to be a normal vector holding the higher dimensional
sphere in a lower dimensional space. Thus if we apply a force to to the cross-section of the
sphere in the plane there should be a force countering it proportional to the normal holding it in
a lower dimensional universe. This counter force would be experienced as inertia. It may even
induce in it an electric field, and we can see how it may do the same equal but opposite for the
electron."
1
t
1
α
2
m
p
h4πr
2
p
Gc
= 6proton s
r
p
t
1
α
m
p
= 1.67262E − 27kg
α
2
=
U
e
m
e
c
2
of 13 35
of 14 35
Since we have!
!
Is a number of protons!
!
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided
by the mass of a proton. But these masses can be considered to cancel and leave pure
number.!
Generating A Table
We make a program that looks for close to whole number solutions so we can create a table of
values for problem solving. I set it at decimal part equal to 0.25. You can choose how may
values for t you want to try, and by what to increment them. Here are the results for
incrementing by 0.25 seconds then 0.05 seconds. Constant to all of this is hydrogen and
carbon. The smaller integer value of seconds gives carbon (6 protons at 1 second) and the
largest integer value of seconds gives hydrogen (1 proton at six seconds) and outside of that
for the other integer values of protons you get are at t>0 and t<1. Equation 1 really has some
interesting properties. Here are two runs of the program( decpart is just me verifying that my
boolean test was working right to sort out whole number solutions):!
By what value would you like to increment?: 0.25!
How many values would you like to calculate for t in equation 1 (no more than 100?): 100!
24.1199 protons 0.250000 seconds 0.119904 decpart !
12.0600 protons 0.500000 seconds 0.059952 decpart !
8.0400 protons 0.750000 seconds 0.039968 decpart !
6.0300 protons 1.000000 seconds 0.029976 decpart !
4.0200 protons 1.500000 seconds 0.019984 decpart !
3.0150 protons 2.000000 seconds 0.014988 decpart !
2.1927 protons 2.750000 seconds 0.192718 decpart !
2.0100 protons 3.000000 seconds 0.009992 decpart !
1.2060 protons 5.000000 seconds 0.205995 decpart !
1.1486 protons 5.250000 seconds 0.148567 decpart !
1.0964 protons 5.500000 seconds 0.096359 decpart !
1.0487 protons 5.750000 seconds 0.048691 decpart !
1.0050 protons 6.000000 seconds 0.004996 decpart !
0.2487 protons 24.250000 seconds 0.248659 decpart !
0.2461 protons 24.500000 seconds 0.246121 decpart !
0.2436 protons 24.750000 seconds 0.243635 decpart !
By what value would you like to increment?: 0.05!
How many values would you like to calculate for t in equation 1 (no more than 100?): 100!
40.1998 protons 0.150000 seconds 0.199837 decpart !
30.1499 protons 0.200000 seconds 0.149879 decpart !
24.1199 protons 0.250000 seconds 0.119904 decpart !
1
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
1
α
2
m
p
h4πr
2
p
Gc
of 15 35
20.0999 protons 0.300000 seconds 0.099918 decpart !
17.2285 protons 0.350000 seconds 0.228500 decpart !
15.0749 protons 0.400000 seconds 0.074938 decpart !
12.0599 protons 0.500000 seconds 0.059950 decpart !
10.0500 protons 0.600000 seconds 0.049958 decpart !
8.0400 protons 0.750000 seconds 0.039966 decpart !
7.0941 protons 0.850000 seconds 0.094088 decpart !
6.0300 protons 1.000000 seconds 0.029975 decpart !
5.2435 protons 1.150000 seconds 0.243457 decpart !
5.0250 protons 1.200000 seconds 0.024980 decpart !
4.1586 protons 1.450000 seconds 0.158605 decpart !
4.0200 protons 1.500000 seconds 0.019985 decpart !
3.1737 protons 1.899999 seconds 0.173673 decpart !
3.0923 protons 1.949999 seconds 0.092296 decpart !
3.0150 protons 1.999999 seconds 0.014989 decpart !
2.2333 protons 2.699999 seconds 0.233325 decpart !
2.1927 protons 2.749999 seconds 0.192719 decpart !
2.1536 protons 2.799999 seconds 0.153564 decpart !
2.1158 protons 2.849998 seconds 0.115782 decpart !
2.0793 protons 2.899998 seconds 0.079303 decpart !
2.0441 protons 2.949998 seconds 0.044061 decpart !
2.0100 protons 2.999998 seconds 0.009993 decpart !
1.2433 protons 4.850000 seconds 0.243294 decpart !
1.2306 protons 4.900001 seconds 0.230607 decpart !
1.2182 protons 4.950001 seconds 0.218177 decpart !
Here is the code for the program:!
#include <stdio.h>!
#include <math.h>!
int main(int argc, const char * argv[]) {!
!
int n;!
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;!
!
do!
{!
printf("By what value would you like to increment?: ");!
scanf("%f", &increment);!
printf("How many values would you like to calculate for t in equation 1 (no more than 100?):
");!
scanf("%i", &n);!
}!
while (n>=101);!
{!
for (int i=0; i<n;i++)!
{!
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));!
!
int intpart=(int)protons[i];!
float decpart=protons[i]-intpart;!
of 16 35
t=t+increment;!
if (decpart<0.25)!
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);!
}}}}!
A very interesting thing here is looking at the values generated by the program, the smallest
integer value, 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest
of the elements heavier than carbon are formed by fractional seconds. These are the
hydrocarbons the backbones biological chemistry.!
Equation 3.2 !
!
!
!
Thus we have six protons (carbon) is one second. We see 1 second predicts the radius of a
proton,…!
In that we have!
!
And the periodic table of the elements is cyclical with 18 groups and!
!
Then perhaps we are supposed to write!
!
In fact, what if the 3 is supposed to be pi, then!
1
6α
2
m
p
h4πr
2
p
Gc
= 1.004996352seconds
1
α
2
m
p
h4πr
2
p
Gc
= (6proton s)(1secon d )
1
α
2
m
p
h4πr
2
p
Gc
= (1proton)(6seconds)
α
2
=
U
e
m
e
c
2
1
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
6 =
1
α
2
m
p
h4πr
2
p
Gc
3
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = 18
of 17 35
!
Then we would say that!
k=18/pi=5.7229577951!
The parameter in our constant with the most uncertainty is the radius of a proton . If the 3 is
supposed to be pi, then the radius of a proton becomes:!
Equation 3.3 !
Which gives!
!
About 95% raw most recent value measured. But, if !
!
Is supposed to be 6 and it is supposed to be multiplied by three to give 18 even which we
need for chemistry so we have 18 protons in the last group of the periodic table which is
important because we need argon with 18 protons for predicting valence numbers of elements
in terms of their need to attain noble gas electron configuration. Then we get!
!
This is in very close agreement with the most recent value measured which is!
!
We now formulate what I call Giordano’s Relationship: Warren Giordano writes in his paper The
Fine Structure Constant And The Gravitational Constant: Keys To The Substance Of The Fabric
Of Space, March 21, 2019:!
In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic
equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where ‘ ’
is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational constant
numerically, but neglecting any units.
Let’s do that!
π
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = 18
r
p
r
p
= kα
2
m
p
Gc
4πh
r
p
= 8.790587E − 16m
1
α
2
m
p
h4πr
2
p
Gc
r
p
= 8.288587E − 16m = 0.829f m
r
p
= 0.833 + / − 0.014
h
1 + α
α
10
23
of 18 35
(6.62607E-34Js)(1+1/137)(1E23)=6.6744E-11 Js!
And it works, G is:!
G=6.67408E-11 N(m2/kg2)!
This is like saying gravitational constant masses equal ten to the 23 Planck velocities:!
!
!
Where m1 is a kilogram and v1 is 1 meter per second. Avogadro’s number is!
!
But, if we are working with kilograms we say!
!
But H=0.001 kg/mol not 1.00 g/mol and carbon is C=0.012 kg/mol not 12.01 g/mol!
Thus if H is the molar mass of hydrogen (1 g/mol=0.001kg/mol) then there are 6E26 atoms in a
kilogram of it. We have!
!
Remembering we now say in this system of units if we are working in
kg.!
But can we say!
!
And our equations become!
!
!
Where E is any element and N is some number. Let us reformulate this as:!
Gm
1
= 10
23
h(1 + α)(v
1
)
Gm
1
= 10
23
h(1 + α)(v
1
)
6E 23atoms/m ol = N
A
N
A
= 6E 26atom s /mole
6 ⋅ secon ds
kilogram s
meter
= 1000h
N
A
G
(1 + α)
N
A
= 6E 26atom s /mol
N
A
H
= 6E 23
6 ⋅ secon ds
kilogram s
2
meter
= h
N
A
H
(1 + α)
G
N ⋅ kilogram s
2
secon ds
meter
= h
N
A
E
(1 + α)
G
of 19 35
Equation 3.4 !
Where and H=1 gram/atom!
Because for hydrogen 1 proton is molar mass 1 gram, for carbon 6 protons is 6 grams and so
on for 6E23 atoms per gram. Thus,…!
!
Since grams and atom cancel we can work in grams even though our equations are in
kilograms. Let us not write H, since formally it is grams per mole of hydrogen but write !
!
We have:!
!
Or,…!
Equation 3.5 !
Where!
!
Let us say we were to consider Any Element say carbon . Then in general!
Equation 3.6 !
We have!
and !
Because there are six grams of protons in carbon which has 6 protons and 6 neutrons and a
molar mass of 12. We have 12-6=6 grams of protons in the 12 grams of protons and neutrons.
Thus!
!
And it follows that!
h
(1 + α)
G
N
A
H = 6.0003
kg
2
⋅ s
m
N
A
= 6.02E23
N
A
H = 6.02E23
atom s
gram
⋅
1gram
atom
= 6.02E23
ℍ = 1
gram
atom
h
(1 + α)
G
⋅ N
A
ℍ = 6.0003kg
2
⋅
s
m
h(1 + α)N
A
ℍ = 6Gx
x = 1.00kg
2
s
m
𝔼
ℂ
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
ℂ =
6gram s
6proton s
N
A
=
6(6E 23proton s)
6gram s
N
A
ℂ = 6E 23
of 20 35
!
We see in general since the atomic number Z is the number of protons in an atom that in
general this holds for all elements because!
!
And,!
!
Therefore we always have:!
Equation 3.7 !
Let us consider then equation 5. We multiply both sides by .!
!
That is we have!
Equation 3.8 !
We will return to this answer, but me we look at the significance of time in Quantum Mechanics
first.!
Thus returning to theorem 1.0 and considering equation 3.1 And 3.4!
Theorem 1.0 The Periodic Table of the Elements employs six fold symmetry. The periods of the
periodic table are periodic over 18 groups, which is 3 times 6, the three being the components
of , the number of values needed to specific a point in such a space.!
Equation 3.1 !
Equation 3.4. !
h
(1 + α)
G
⋅ N
A
ℂ = 6.0003kg
2
⋅
s
m
𝔼
N
A
=
Z ⋅ 6E 23protons
Z ⋅ grams
𝔼 =
Z ⋅ grams
Z ⋅ proton s
N
A
⋅ 𝔼 = 6E 23
r
p
m
2
p
r
p
m
2
p
= 6.000kg
2
s
m
⋅
0.833E − 15m
(1.67262E − 27kg)
2
= 1.7865E 39s
r
p
m
3
p
⋅ h
(1 + α)
Gx
⋅ N
A
𝔼 = 1.7865E 39s
ℝ
3
1
t
1
α
2
m
p
h4πr
2
p
Gc
= 6proton s
h
(1 + α)
G
N
A
H = 6.000
kg
2
⋅ s
m
of 21 35
Thus in 3.1 we have 6 protons and in 3.4 we have . Equation 1 is determined by
. !
Buckminster Fuller writes in Synergetics:!
205.01 The geometrical model of energy configurations in synergetics is developed from a
symmetrical cluster of spheres, in which each sphere is a model of a field of energy all of whose
forces tend to coordinate themselves shuntingly or pulsatively, and only momentarily in
positive or negative asymmetrical patterns relative to, but never congruent with, the eternality of
the vector equilibrium…This closest packing of spheres in 60-degree angular relationships
demonstrates a finite system in Universal geometry.
6
kg
2
⋅ s
m
t
1
= 1s
of 22 35
"
of 23 35
Thus we see carbon is 8 tetrahedra and 6 octahedra, each of which is 2 tetrahedra I found
them precisely this in their two dimensional analogs the regular dodecahedron and regular
octagon:!
We see that the atomic radius of silicon the core element of artificial intelligence (transistor
technology) fits together with the core element of biological life carbon if the silicon is taken as
inscribed in a regular dodecagon and the carbon is taken as inscribed in a regular octagon. We
have:!
, , !
, , !
Apothem: !
For a regular dodecagon:!
=!
The radius of a silicon atom is Si=0.118nm and that of carbon is
C=0.077nm:!
!
!
!
This has an accuracy %!
But, we still have the question of 1 second in equation 3.1:!
!
We see that the vector equilibrium employs 60-degree coordination. And that it involves a layer
of 12 spheres. However we have a 12 hour daytime and and 12 hour night time, which is a 24
hour day (Earth rotates through 15 deg an hour, dodecagon). But there are 60 minutes in an
hour and 60-degree coordination is connected to that in that the earth rotates through 360
degrees in 24 hours and 360/60 degrees is our six-fold symmetry of the periodic table. Each
hour then of the 24 hours is divided by 60 minutes and each minute divided by 60 is 1 second.
D = 1 + 2x
x
2
+ x
2
= 1
2
2x
2
= 1
2x
2
= 1
x = 2/2
D = 1 + 2
a = (1 + 2)/2 = 1.2071
a =
s/2
tan(θ/2)
=
0.5
ta n15
∘
(15
∘
)
Si
C
=
0.118
0.077
= 1.532
a
S
i
a
C
=
1.866
1.2071
= 1.54585
1.532
1.54585
100 = 99
1
t
1
α
2
m
p
h4πr
2
p
Gc
= 6proton s
of 24 35
We have our one second. Indeed I find the size of the solar system is connected to Silicon
through its counterpart as a semiconductor material germanium. Let’s look at that:!
While we have!
!
Is a number of protons!
!
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided
by the mass of a proton. But these masses can be considered to cancel and leave pure
number. We have that!
!
!
Interestingly 78% is the percent of N2 in the atmosphere and 21% is the percent of O2 in the
atmosphere (by volume). These are the primary constituents that make it up. The rest is
primarily argon and CO2. This gives the molar mass of air as a mixture is:!
!
Now interestingly, I have found this connected to the solar system by considering a mixture of
silicon, the primary constituent of the Earth crust, and germanium just below it in the periodic
table, in the same proportions of 78% and 21%. Silicon is also the primary second generation
semiconductor material (what we use today) and germanium is the primary first generation
semiconductor material (what we used first). The silicon is directly below our carbon of one
proton-second, silicon directly below that, and germanium directly below that, in the periodic
table. So we are moving directly down the periodic table in group 14. The density of silicon is
2.33 g/cm3 and that of germanium is 5.323 g/cm3. Let us weight these densities with our 0.21
and 0.78:!
!
Now consider this the starting point for density of a thin disc decreasing linearly from the Sun
to Pluto (49.5AU=7.4E14cm). Thus,…!
!
1
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
1
α
2
m
p
h4πr
2
p
Gc
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.78
∫
3
2
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
−
π − 4
2 2
= 0.21
0.78N 2 + 0.21O2 = 29.0g /mol
0.21Si + 0.78Ge = 4.64124g/cm
3
ρ(r) = ρ
0
(
1 −
r
R
)
of 25 35
Thus,…!
!
Or,…!
!
Thus,…!
!
If we add up the masses of the planets in our solar system they are 2.668E30 grams.!
Since!
!
Meaning mixing germanium and silicon in the same proportion that occurs with N2 and O2 in
the atmosphere and with!
!
!
Where!
!
In the first integral. See the following pages to see the computation of the mass of the planets
in the solar system…!
!
As we can see Jupiter carries the majority of the mass, Saturn a pretty large piece, and
somewhat large Uranus and Neptune. We don’t even list Pluto’s mass. When we consider!
M =
∫
2π
0
∫
R
0
ρ
0
(
1 −
r
R
)
rdrdθ
M =
πρ
0
R
2
3
M =
π(4.64124)(7.4E14)
2
3
= 2.661E 30
2.661
2.668
(100) = 99.736
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.78
∫
3
2
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
−
π − 4
2 2
= 0.21
6 =
1
α
2
m
p
h4πr
2
p
Gc
of 26 35
!
!
We are considering and . These come from!
!
!
From 30 degrees to 45 degrees is 15 degrees. The Earth rotates through 360/24 is 15 degrees
per hour. The hour is divided into 60 minutes and minutes into 60 seconds…!
We have said!
!
For a thin disc. Thus we have a definition for the radius of the solar system, :!
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.78
∫
3
2
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
−
π − 4
2 2
= 0.21
2
3
2 cos(45
∘
) = 2
2 cos(30
∘
) = 3
M =
πρ
0
R
2
3
R
s
of 27 35
Equation 3.9 !
Where!
Equation 3.10 !
Equation 3.111 !
Equation 3.12 !
Equation 3.13 !
is the mass of all the planets. We also have that the molar mass of air to the molar mass of
water is approximately the golden ratio. The interesting thing is we determine a definition for
the radius of the solar system and predict the hydrocarbons (backbones of life) all in one fell
swoop. Thus since if the universe is mostly hydrogen and helium if we can predict their relative
abundances with proton-seconds then we can feel quite certain we are formulating things
correctly, consider the following gaussian distribution illustrated…"
R
s
=
3M
p
π(0.78Ge + 0.21Si )
1
α
2
m
p
h4πr
2
p
Gc
∫
3
2
1
t
2
dt = 0.78
∫
3
2
cos
−1
(x /2)d x = 0.21
air = 0.78N
2
+ 0.21O
2
air
H
2
O
≈ Φ
M
p
of 28 35
"
of 29 35
We consider a Gaussian wave-packet at t=0:!
!
We say that d which in quantum mechanics would be the delocalization length when squared is
. A is the amplitude and we might say it is . We write the wave packet as a
Fourier transform which is:!
!
We use the identity that gives the integral of a quadratic:!
!
Solve the equation!
!
With the initial condition!
!
A plane wave is the solution:!
!
Where, !
The wave-packet evolves with time as!
!
Calculate the Gaussian integral of dp!
and !
The solution is:!
ψ (x,0) = Ae
−
x
2
2d
2
Si − C
C
Ga − P
C
ψ (x,0) = Ae
−
x
2
2d
2
=
∫
dp
2π ℏ
ϕ
p
e
i
h
px
∫
∞
−∞
e
−α
2
x+βx
d x =
π
α
e
β
2
4α
iℏ∂
t
ψ (x, t) =
p
2m
ψ (x, t)
ψ (x,0) =
∫
dp ⋅ e
p
2
d
2
2ℏ
2
⋅ e
−
i
h
px
e
i
ℏ
( px−ϵ( p)t)
ϵ( p) =
p
2
2m
ψ (x, t) =
∫
dp ⋅ e
p
2
d
2
2ℏ
2
⋅ e
−
i
h
( px−
p
2
2m
t)
α =
d
2
2ℏ
2
+
it
2mℏ
β =
i x
ℏ
of 30 35
where !
We notice here one of the things you can do with equation 4 is integrate from 0.5 sec to 1 sec
and you get one which multiplied by the constant which is six yields six. Now look up 0.5
seconds from the data output from the program and it is magnesium, then go to one second
and it is carbon, thus the integral from magnesium in time to carbon in time is carbon in
protons. Now consider life as we know it is based on carbon because it has four valence
electrons, but it is not based on silicon, which has four valence electrons as well, because in
the presence of oxygen it readily forms SiO2 (sand or glass) leaving it unavailable to nitrogen,
phosphorus, and hydrogen to make make amino acids the building blocks of life. But silicon
can be doped with phosphorus, boron, gallium, and arsenic to make semiconductors --
transistor technology from which we can build artificial life (artificial intelligence, AI). We can
integrate over many time ranges to explore millions of more facets to the equation:!
!
!
!
We should say!
or !
The way I am using equation 1 is . We have:!
Equation 3.14!
!
ψ
2
= exp
[
−
x
2
d
2
⋅
1
1 + t
2
/τ
2
]
τ =
m d
2
ℏ
1
α
2
m
p
h4πr
2
p
Gc
∫
t
C
t
Mg
1
t
2
dt = 6 = carbon(C )
t
Mg
= 0.5secon ds
t
C
= 1secon d
d =
Si − C
C
=
4
3
d
2
=
16
9
τ = d
2
ψ
2
= exp −
C
2
x
2
(Si − C )
2
⋅
1
1 +
[
ℏC
2
m(Si − C)
2
]
2
t
2
of 31 35
Thus for hydrogen:!
!
=74%.!
For Helium:!
=26%!
This is interesting because the Universe is about 74% Hydrogen and 24% Helium, the rest of
the elements making up the other 2%. Thus we can say or . We have
multiplied the first by 1 for Hydrogen element 1 (1 proton), and the second by 2 for helium
element 2 (2 protons). In a sense then, the probabilities represent the probability of finding
hydrogen and helium in the Universe. Hydrogen much of the helium were made theoretically in
the Big Bang of the big bang theory. The other elements were synthesized from these by the
stars. The difference between hydrogen and helium is that some of the helium was made in
stars, that may be why it is 26% not 24% because it could include the remaining 2% of
elements in the universe. It must be kept in mind the data we have on the universe as a whole
for relative abundances can only currently be ballpark figures."
ψ
2
= exp −
9
16
x
2
⋅
1
1 +
ℏ
2
81
m
2
256
t
2
ψ
2
= (1)exp −
9
16
(1proton)
2
⋅
1
1 +
(0.075)81
(1)256
(6secon ds)
2
ψ
2
= (2)exp −
9
16
(2proton)
2
⋅
1
1 +
(0.075)81
(4)256
(3secon ds)
2
ℏ
2
= 0.075
ℏ = 0.27386
of 32 35
4.0 Formulation of Proton-Seconds
We can actually formulate this differently than we have. We had!
!
!
But if t1 is not necessarily 1 second, and t6 is not necessarily six seconds, but rather t1 and t2
are lower and upper limits in an integral, then we have:!
Equation 4.1 !
This Equation is the generalized equation we can use for solving problems.!
Essentially we can rigorously formulate the notion of proton-seconds by considering!
Equation 4.2 !
Is protons-seconds squared where current density is and ( can also be
). We say!
Equation 4.3. !
Keeping in mind q is not charge (coulombs) but a number of charges times seconds, here a
number of protons. It is!
Equation 4.4 !
Dividing Equation 4.2 through by by t:!
Equation 4.5 !
!
Which is proton-seconds. Dividing through by t again:!
1
t
1
1
α
2
m
p
h4πr
2
p
Gc
= 6proton s
1
t
6
1
α
2
m
p
h4πr
2
p
Gc
= 1proton
1
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
∫
t
qdt = t
2
∬
S
ρ(x, y, z)d x d y
J = ρ
v
ρ = Q /m
3
ρ
Q /m
2
Q =
∫
V
ρdV
ℕ =
1
α
2
m
p
h4πr
2
p
Gc
1
α
2
m
p
h4πr
2
p
Gc
∫
t
dt
t
= t
∬
S
ρ(x, y, z)d x d y
of 33 35
Equation 4.6 !
We see that if where and then J is I/m2 (current per square
meter) is analogous to amperes per per square meter which are coulombs per second through
a surface. Thus we are looking at a number of protons per second through a surface. Thus we
write:!
!
Is carbon where 0.5 seconds is magnesium (Mg) from the values of time corresponding to
protons in the output from our program and 1.0 seconds is carbon (C). We see we have the
following theorem:!
Equation 4.7 !
So as an example,…!
!
Is fluorine (F). Divide by xy with x=y=1 and we have current density. And multiply by 1 second
which is carbon and we have protons per square meter. Charge conservation gives:!
!
Is 9-9=0. In general…!
!
!
"
1
α
2
m
p
h4πr
2
p
Gc
∫
t
dt
t
2
= proton s
J = ρ
v
ρ = Q /m
3
v = m /s
1
α
2
m
p
h4πr
2
p
Gc
∫
t
C
t
Mg
dt
t
2
= 6
∫
1.0
0.5
t
−2
dt = − 6(1 − 2) = 6
1
α
2
m
p
h4πr
2
p
Gc
∫
t
dt
t
3
=
∬
S
J ⋅ d
S
1
α
2
m
p
h4πr
2
p
Gc
∫
1.0
0.5
dt
t
3
=
∬
S
J ⋅ d
S = − 3
(
1 −
1
0.25
)
= 9
proton s
secon d
∂ρ
∂ t
+ ∇ ⋅
J = 0
J (x, y, z) = (0,0,J ) = − J
k
d
S = d xdy
k
J ⋅ d
S = (0,0,J ) ⋅ (0,0,d x d y) = − Jd xdy
of 34 35
!
We consider the surface through which J passes a circle of radius R where J decreases linearly
as r with at the center we have:!
Equation 4.8 !
!
!
J
0
∬
S
J ⋅ d
S = J
0
∫
2π
0
∫
R
0
(
1 −
r
R
)
rdrdθ =
π J
0
R
2
3
of 35 35
The Author!
